## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{\theta\to0}\frac{\sin(\sin\theta)}{\theta}=1$$
$$A=\lim_{\theta\to0}\frac{\sin(\sin\theta)}{\theta}$$ $$A=\lim_{\theta\to0}\frac{\sin(\sin\theta)}{\sin\theta}\times\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ We have $$\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$$ For $\lim_{\theta\to0}\frac{\sin(\sin\theta)}{\sin\theta}$, take $\sin\theta=x$, then as $\theta\to0$, we have $x=\sin\theta\to\sin0=0$ So $$\lim_{\theta\to0}\frac{\sin(\sin\theta)}{\sin\theta}=\lim_{x\to0}\frac{\sin x}{x}=1$$ Therefore, $$A=1\times1=1$$